Average Velocity and Instantaneous Velocity

The average velocity of a particle in a time interval t1 to t2 is defined as its displacement divided by the time interval.

For example – If the particle is at a point A at time t = t1 and at B at time t = t2
The displacement in this time interval is the vector AB.

Then average velocity in this time interval is

\begin{equation} \vec{v}_{a v}=\frac{\overrightarrow{A B}}{t_{2}-t_{1}} \end{equation}

Like displacement, average velocity is also a Vector quantity.

Moreover we can further simplify this equation as from triangle AOB it’s clear that

\begin{equation} \vec{AB} = \vec{AO} + \vec{OB} \\ \vec{AO} = – \vec{OA} \\ \implies \vec{AB} = vec{OB} – \vec{OA} \\ \vec{OB} = \vec{r_{2}} \\ \vec{OA} = \vec{r_{1}} \\ \implies \vec{AB} = \vec{r_{2}} – \vec{r_{1}} \\ \text{Thus } \vec{v}_{a v}=\frac{\overrightarrow{A B}}{t_{2}-t_{1}} \text{ can be rewritten as } \\ \vec{v}_{a v}= \frac {\vec{r_{2}} – \vec{r_{1}}} {t_{2} – t_{1}} \end{equation}

It’s clear from above equation that average velocity takes into consideration only final and initial positions of object and doesn’t care about how object went from initial to final position.

Let’s understand this using an example problem.

Problem
A table clock has it minute hand 4.0 cm long.
Find the average velocity of the tip of the minute hand.

1) Between 6:00 am to 6:30 am
2) Between 6:00 am to 6:30 pm
Solution
1) Between 6:00 am to 6:30 am

At 6:00 am
minute hand of clock is at 12 mark on clock

At 6:30 am
minute hand of clock is at 6 mark on clock
Displacement = 8.0 cm
Time Taken = (from 6:00 am to 6:30 am is 30 minutes) = 30 min = 30 × 60 = 1800 s
Time Taken = 1800 s

Average velocity = vav = Displacement/Time Taken = 8.0 / 1800 = 4.4 × 10-3 cm/s
Solution
2) Between 6:00 am to 6:30 pm

At 6:00 am
minute hand of clock is at 12 mark on clock

At 6:30 am
minute hand of clock is at 6 mark on clock

Similar to above question, displacement of minute hand in this case will be 8.0 cm
Displacement = 8.0 cm

But in this case time has gone from 6:00 am to 6:30 pm which is 12 hours 30 minutes which is 45000 seconds.

Displacement = 8.0 cm
Time Taken = 45000 s

Average velocity = vav = Displacement/Time Taken = 8.0 / 45000
= 1.8 × 10-4 cm/s

⇒ Average velocity = vav = 1.8 × 10-4 cm/s

Let’s now discuss Instantaneous Velocity

Average Velocity of the particle in a short time interval t to t + Δt is defined as follows.

\begin{equation} \vec{v_{av}} = \frac {Δ\vec{r}} {Δt} \end{equation}

where Δr vector is the displacement in the time interval Δt

Now if we make Δt vanishingly small and find the limiting value of Δr/Δt
This value is Instantaneous Velocity of the particle at time t.

\begin{equation} \vec{v}=\lim _{\Delta t \rightarrow 0} \frac{\Delta \vec{r}}{\Delta t}=\frac{d \vec{r}}{d t} \end{equation}

Let’s understand Instantaneous Velocity using an example.

Problem
Let’s suppose that a particle is moving on a path and it’s displacement function is given as 5t2.
Then find out Instantaneous Velocity of this particle at t = 10 s
Solution
Instantaneous Velocity at a particular instance of time t is given as dr/dt
r = 5t2 as per question

dr/dt = d(5t2)/dt = 10t

Thus Instantaneous Velocity function of particle is 10t

At t = 10 s
Instantaneous Velocity = 100 m/s
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